I have 2 classes, with many-to-many relationship, my goal is to fill an 'item' list with data from that 2 models, here are my models:
class Bakery(models.Model):
title = models.CharField('restaurant_name', max_length=100)
class DeliveryService(models.Model):
title = models.CharField('deliveryservice_name', max_length=100)
bakery = models.ManyToManyField(Bakery)
Here is the logic on my 'views' file:
item = []
bakerys = Bakery.objects.all()
for i in bakerys:
item.append(i.title)
item.append(i.deliveryservice.title)
I hope you got what exactly I want to accomplish. My current 'views' file logic is wrong and I know it, I just does not know what can I do to solver this problem. Thank you for your time.
The following seems to do what you're asking for. But it seems odd that you want to create a list with all the titles for different objects all mixed together and likely have duplicates (if a delivery service is linked to more than one bakery it'll be added twice).
item = []
bakerys = Bakery.objects.all()
for i in bakerys:
item.append(i.title)
for j in i.deliveryservice_set.all():
item.append(j.title)
You should really read up on the many-to-many functionality of the ORM. The documentation is pretty clear on how to do these things.
Sayse had a good answer too if you really just want all the titles. Their answer also groups everything in tuples and accomplishes it with more efficiency by using fewer db queries. Their answer was: Bakery.objects.values('title', 'deliveryservice__title')
Related
I have a data model like this:
class Post(models.Model)
name = models.CharField(max_length=255)
class Tag(models.Model)
name = models.CharField(max_length=255)
rating = models.FloatField(max_length=255)
parent = models.ForeignKey(Post, related_name="tags")
I want to get Posts that have a tag, and order them by the tags rating.
something like:
Posts.objects.filter(tags__name="exampletag").order_by("tags(name=exampletag)__rating")
Currently, I am thinking it makes sense to do something like
tags = Tags.objects.filter(name="sometagname").order_by("rating")[0:10]
posts = [t.parent for t in tags]
But I like to know if there is a better way, preferably querying Post, and getting me back a queryset.
Edit:
I don't think this: (Edit 2 - this does give the correct sorting!)
Posts.objects.filter(tags__name="exampletag").order_by("tags__rating")
will give the correct sorting, as it does not sort only by the related item with name "exampletag"
Something like the following would be needed
Posts.objects.filter(tags__name="exampletag").order_by("tags(name=exampletag)__rating")
I've been looking over the django docs, and it seem "annotate" nearly works - but I don't see a way to use it to select a tag by name.
Edit 2
Both the Answers are correct! See my comments to observe some epic brain-farts (one test, the results WERE in order, the other i filter and sort by different tags!)
how it works
the query
Posts.objects.filter(tags__name="exampletag").order_by("tags__rating")
and
Posts.objects.filter(tags__name="exampletag").filter(tags__name="someothertag").order_by("tags__rating")
will work correctly and by sorted by the rating of "exampletag"
it seems the tag(From a ForeignKey BackReference Set) used for sorting when calling order_by is the one in the first filter.
You can do like:
tags = Tags.objects.filter(name="sometagname")
posts = Post.objects.filter(tags__in=tags).order_by('tags__rating')
Even shorter than Anush's, with a JOIN rather than a subquery:
Post.objects.filter(tags__name='exampletag').order_by('tags__rating')
I'm trying to extract free-text answer submitted by workers of Amazon Mechanical Turk using the boto library.
assignments = conn.get_assignments(hit_id)
for assignment in assignments:
worker = assignment.WorkerId
answer = assignment.Answer
Here I expect answer to be a free-text string (the only thing that the HIT asks workers to submit) submitted by a worker, however, the code above doesn't give me that. What am I missing here?
In boto in order to get the FreeText information you are looking for, you'll need to iterate over the assignment property answers. Unless you have submitted multiple forms, your form should be the first index.
This list is of type QuestionFormAnswer
Here is boto documentation on QuestionFormAnswer
http://sourcecodebrowser.com/python-boto/2.3.0/classboto_1_1mturk_1_1connection_1_1_question_form_answer.html
You can see that the properties you actually want are qid and fields
Here is some updated code that should make better sense.
assignments = conn.get_assignments(hit_id)
for assignment in assignments:
worker_id = assignment.WorkerId
# Iterate through question forms answers which are our fields
for question_form_answer in assignment.answers[0]:
field_id = question_form_answer.qid
field_value = question_form_answer.fields
I think the assignment object in the above example will have an attribute called answers which is a list of QuestionFormAnswer objects. Each of these objects should have an attribute called FreeText.
I have two files that I want to compare with each other and form a list. Each file have their own class. Book and Person. In these, I have different attributes. The ones I want to compare are: person.personalcode == book.borrowed. From this I want a list of all the borrowed books. I have started like this:
for person in person_list:
for book in booklibrary_list:
if person.personalcode == book.borrowed:
person.books.append(book, person)
for person in person_list:
if len(person.books) > 0:
print(person.personalcode + "," + person.firstname + person.lastname + "have borrowed the following books: ")
for book in person.books:
print(book)
for person in person_list:
person.books = []
But it does not work, what have I missed or done wrong?
Posting as an answer as this is too long for a comment.
First: improve your question. Show how you construct the Person and the Book class, and how you populate them. Describe what the personalcode is and how come personalcode would be the same as a book code. Some sample data and a bit more code would make this easier to answer.
Second: reading your other question, you seem to be storing your data in a text file, loading and querying, modifying and saving the data directly. This will lead you to problems and instead you should consider going down one of two lines:
Use an SQL database, possibly the easiest to start with is SQLite as it does not need a server to be set up and there is a module in the standard library that is very easy to use. Store your data there and you will find it easier in the long run.
Use Python objects (e.g. three classes: Person, Book, and BorrowedBook), manage lists of them within the program, and use shelve from the standard library to store and retrieve these lists of objects between queries.
The use of shelve would be easier if you have not used SQL before, and I hope you will forgive the pun when I say that it might be very appropriate for a book-related application!
I have a model similar to this one:
class MyModel(models.Model):
name = models.CharField(max_length = 30)
a = models.ForeignKey(External)
b = models.ForeignKey(External, related_name='MyModels_a')
def __unicode__(self):
return self.a + self.b.name + self.b.name
So when I query it I get something like this:
>>> MyModel.objects.all()
[<MyModel: Name1AB>,<MyModel: Name2AC>,<MyModel: Name3CB>,<MyModel: Name4BA>,<MyModel: Name5BA>]
And I'd like to represent this data similar to the following.
[[ [] , [Name1AB] , [Name2AC] ]
[ [Name4BA, Name5BA] , [] , [] ]
[ [] , [Name3CB] , [] ]]
As you can see the rows would be 'a' in the model; and the columns would be 'b'
I can do this, but it takes a long of time because in the real database I have a lot of data. I'd like to know if there's a Django built in way to do this.
I'm doing it like this:
mymodel_list = MyModel.objects.all()
external_list = External.objects.all()
for i in external_list:
for j in external_list:
print(mymodel_list.filter(a=i).filter(arrl=j).all(),end='')
print()
Thanks
Three ways of doing it but you will have to research a bit more. The third option one may be the most suitable for what you are looking for.
1) Django queries
The reason it is taking a long time is because you are constantly accessing the database in this line:
print(mymodel_list.filter(a=i).filter(arrl=j).all(),end='')
You may have to start reading what the Django documentation say about the way of working with queries. For what you are doing you have to create the algorithm to avoid the filters. Using MyModel.objects.order_by('a') may help you to build a efficient algorithm.
2) {% ifchanged ...%} tag
I suppose you are using print to post your answer but you probably need it in html. In that case, you may want to read about the ifchanged tag. It will allow you to build your matrix in html with just one db access.
3) Many to many relations
It seems you are sort of simulating a many to many relation in a very peculiar way. Django has support for many to many relations. You will need an extra field, so you will also have to read this.
Finally, for doing what you are trying with just one access to the database, you will need to read prefetch_related
There's no built-in way (because what you need is not common). You should write it manually, but I'd recommend retrieving full dataset (or at least dataset for one table line) and process it in Python instead of hitting DB in each table cell.
I'm trying to figure out if there's a way to do a somewhat-complex aggregation in Django using its ORM, or if I'm going to have to use extra() to stick in some raw SQL.
Here are my object models (stripped to show just the essentials):
class Submission(Models.model)
favorite_of = models.ManyToManyField(User, related_name="favorite_submissions")
class Response(Models.model)
submission = models.ForeignKey(Submission)
voted_up_by = models.ManyToManyField(User, related_name="voted_up_responses")
What I want to do is sum all the votes for a given submission: that is, all of the votes for any of its responses, and then also including the number of people who marked the submission as a favorite.
I have the first part working using the following code; this returns the total votes for all responses of each submission:
submission_list = Response.objects\
.values('submission')\
.annotate(votes=Count('voted_up_by'))\
.filter(votes__gt=0)\
.order_by('-votes')[:TOP_NUM]
(So after getting the vote total, I sort in descending order and return the top TOP_NUM submissions, to get a "best of" listing.)
That part works. Is there any way you can suggest to include the number of people who have favorited each submission in its votes? (I'd prefer to avoid extra() for portability, but I'm thinking it may be necessary, and I'm willing to use it.)
EDIT: I realized after reading the suggestions below that I should have been clearer in my description of the problem. The ideal solution would be one that allowed me to sort by total votes (the sum of voted_up_by and favorited) and then pick just the top few, all within the database. If that's not possible then I'm willing to load a few of the fields of each response and do the processing in Python; but since I'll be dealing with 100,000+ records, it'd be nice to avoid that overhead. (Also, to Adam and Dmitry: I'm sorry for the delay in responding!)
One possibility would be to re-arrange your current query slightly. What if you tried something like the following:
submission_list = Response.objects\
.annotate(votes=Count('voted_up_by'))\
.filter(votes__gt=0)\
.order_by('-votes')[:TOP_NUM]
submission_list.query.group_by = ['submission_id']
This will return a queryset of Response objects (objects with the same Submission will be lumped together). In order to access the related submission and/or the favorite_of list/count, you have two options:
num_votes = submission_list[0].votes
submission = submission_list[0].submission
num_favorite = submission.favorite_of.count()
or...
submissions = []
for response in submission_list:
submission = response.submission
submission.votes = response.votes
submissions.append(submission)
num_votes = submissions[0].votes
submission = submissions[0]
num_favorite = submission.favorite_of.count()
Basically the first option has the benefit of still being a queryset, but you have to be sure to access the submission object in order to get any info about the submission (since each object in the queryset is technically a Response). The second option has the benefit of being a list of the submissions with both the favorite_of list as well as the votes, but it is no longer a queryset (so be sure you don't need to alter the query anymore afterwards).
You can count favorites in another query like
favorite_list = Submission.objects.annotate(favorites=Count(favorite_of))
After that you add the values from two lists:
total_votes = {}
for item in submission_list:
total_votes[item.submission.id] = item.voted_by
for item in favorite_list:
has_votes = total_votes.get(item.id, 0)
total_votes[item.id] = has_votes + item.favorites
I am using ids in the dictionary because Submission objects will not be identical. If you need the Submissions themselves, you may use one more dictionary or store tuple (submission, votes) instead of just votes.
Added: this solution is better than the previous because you have only two DB requests.